What is the vertex form of $y= 4t^2-12t+8$ ?

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Answer 1 · 2017-08-18T15:25:45

$y = 4(t-3/2)^2 -1$

Vertex form is given as $y = a(x+b)^2+c$ ,

where the vertex is at $(-b,c)$

$y = 4t^2 -12t +8$

$y = 4(t^2 -color(blue)(3)t +2)" "larr$ take out the factor of $4$

$y = 4(t^2 -3t color(blue)( +(3/2)^2 -(3/2)^2) +2)$

$[color(blue)(+(3/2)^2 -(3/2)^2=0)]" "larr +(b/2)^2 -(b/2)^2$

$y = 4(color(red)(t^2 -3t +(3/2)^2) color(forestgreen)( -(3/2)^2 +2))$

$y = 4(color(red)((t-3/2)^2) color(forestgreen)( -9/4 +2))$

$y = 4(color(red)((t-3/2)^2) color(forestgreen)( -1/4))$

Now distribute the $4$ into the bracket.

$y = color(red)(4(t-3/2)^2) +color(forestgreen)(4( -1/4))$

$y = 4(t-3/2)^2 -1$

What is the vertex form of y= 4t^2-12t+8 y= 4t^2-12t+8 ?